# If the Brewster's angle is considered to happen with no reflection, then how is the refraction angle considered to be 90°?

So basically my questions are these:

1. Is the refraction angle always 90 in relation to Brewster's angle?
2. And if the refraction angle is the angle between the reflected ray and the ray that passed the surface, and if there's no reflection for the Brewster's angle(which is how Wikipedia defined it), then how is there a refraction angle?

Giving many thanks to all those who will probably answer this in advance 😊 So English is my second language and I would really appreciate it if you kept the words simple.

• Brewster's angle affects only the light which is polarized along the plane of incidence of the light (p-polarized). The light polarized in the orthogonal direction (s-polarized) refracts and reflects as usual. – José Andrade Feb 16 at 13:41
• You may be confusing Brewster angle with critical angle for total internal reflectin. – nasu Feb 16 at 14:16
• @nasu its clear he is talking about Brewster's angle...he even mentions the wikipedia page. I think the confusion lies on the assumption that the no reflection means for all light and not solely for p-polarized light. – José Andrade Feb 17 at 10:14

The angle of refraction is not 90$${}^\circ$$. The angle between the refracted ray and what would be the reflected ray is 90$${}^\circ$$. The reason is that the source of reflected light is the polarization of the medium. The direction of the reflected ray is fixed, determined by the direction of the incident ray. Brewster' angle is that incident angle that produces polarization of the medium that's in the same direction as the reflected ray. An oscillating polarization can not generate light that propagates in the direction of polarization. The direction of polarization is perpendicular to the direction of propagation. So there can be no reflection.
At Brewster's angle there is still reflection of TE polarised light. Also, by Snell's law the angle of reflection equals the angle of incidence (+$$\pi$$).